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Popular >

x^2y^'-y=0

Solution

x^{2} y^{^{'}} - y = 0

Solution

y = e^{- \frac{1}{x} + c_{1}}

Solution steps

x^{2} y^{'} (x) - y = 0

Solve separable ODE:

y = e^{- \frac{1}{x} + c_{1}}

y = e^{- \frac{1}{x} + c_{1}}

Popular Examples

(dy)/(dx)+((1+y^3)x)/(y^2(1+x^2))=0

\frac{d y}{d x} + \frac{( 1 + y ^{3} ) x}{y ^{2} ( 1 + x ^{2} )} = 0

(2y+3)dy=(x^2+6x)dx

(2 y + 3) d y = (x^{2} + 6 x) d x

(dy)/(dt)=1+t+y^2+ty^2

\frac{d y}{d t} = 1 + t + y^{2} + t y^{2}

2y+3+(2x-4)(dy)/(dx)=0

2 y + 3 + (2 x - 4) \frac{d y}{d x} = 0

(x-1)((dy)/(dx))=(4-y^2)

(x - 1) (\frac{d y}{d x}) = (4 - y^{2})

Frequently Asked Questions (FAQ)

What is the solution for x^2y^'-y=0 ?
The solution for x^2y^'-y=0 is y=e^{-1/x+c_{1}}